Free Energy & Pressure
• Remember that pressure computations must only
be pertaining to GASES!
• If the partial pressures are known, you can use
this formula to solve problems…
∆G = ∆G o + RT ln Q
Where ∆G o = free energy from calculations
(∆G o = ∆H o - T ∆S o (beware of units)
R = 8.31 J/K•mol
T = temp. in K
And Q = reaction quotient
The Reaction Quotient = Q
Q = Partial pressures Products raised to coeff. Power
Partial pressures Reactants raised to coeff. Power
•
•
Let’s try one –use appendix 4 in text
Calculate ∆G for this reaction at these
conditions… 2 NO2(g)
N2O4(g)
(T =298 K, pNO2 = .35 atm, pN2O4 = .50 atm)
1. Find ∆G0 first…
∆G0 = (∆
∆Gf0 N2O4(g)-(∆
∆Gf0 2NO2(g))
∆G0 = (98 kJ) – (2(52 kJ)) = -6.0 kJ
Problem, continued
∆G0 = -6.0 kJ rxn
2. Find Q =[pN2O4] = (.50 atm) = 4.08
[pNO2]2
(.35 atm)2
Finish ∆G = ∆G0 + RT ln Q
∆G=(-6.0 kJ) +(8.31 x 10-3kJ/K)(298K)ln(4.08)
∆G = (- 6.0 kJ) + (3.48 kJ) = -2.52 kJ
1
Thermodynamics and Keq
FACT: ∆Gorxn is the change in free energy
when pure reactants convert COMPLETELY to
pure products.
products.
FACT: ProductProduct-favored systems have …
Keq >1
Therefore, both ∆G˚rxn and Keq are
related to reaction favorability.
∆G, ∆G˚, and Keq
• ∆G is change in free energy at nonnon-standard
conditions.
• ∆G is related to ∆G˚
• ∆G = ∆G˚ + RT ln Q
where Q = reaction quotient
• When Q < K reaction is spontaneous
• Q > K reaction is NOT spontaneous
• When Q = K reaction is at equilibrium
• When ∆G = 0 reaction is at equilibrium
• Therefore, ∆G˚ = - RT ln K
Thermodynamics and Keq
Keq is related to reaction favorability
and so to ∆Gorxn.
The larger the value of K the more
negative the value of ∆Gorxn
∆Gorxn = - RT lnK
where R = 8.31 J/K•mol or
8.31 x 10-3 kJ/K • mol
2
Thermodynamics and Keq
∆Gorxn = - RT lnK
Calculate K for the reaction at 298 K
N2O4 ----->2
>2 NO2
∆Gorxn = +4.8 kJ/mol
∆Gorxn = +4800 J/mol = - (8.31 J/Kmol)(298 K) ln K
4800 J
lnK = = - 1.94
(8.31 J/K)(298K)
K = 0.14
When ∆Gorxn > 0, then K < 1 (reactant(reactant-favored)
∆G, ∆G˚, and Keq
Product--favored reaction.
Product
2 NO2 --->
---> N2O4
o
∆G rxn = – 4.8 kJ
Here ∆Grxn is less than
∆Gorxn , so the state with
both reactants and
products present is more
stable than complete
conversion..
conversion
∆G, ∆G˚, and Keq
Reactant-favored reaction.
ReactantN2O4 ----->2
>2 NO2
∆Gorxn =
+4.8 kJ
Here ∆Gorxn is greater than
∆Grxn , so the state with
both reactants and products
present is more stable than
complete conversion.
3
Thermodynamics and Keq
Keq is related to reaction favorability.
When ∆Gorxn < 0, reaction moves
energetically “downhill”
∆Gorxn is the change in free energy when
reactants convert COMPLETELY to
products.
4